0
$\begingroup$

QUESTION(see image attached): enter image description here

Really need help here can someone help me show how we get the solution as shown in the image file attached above.

okay since I was asked by one of the users commenting asking me to show what I had done; please see attached image file below

Thank You.

enter image description here

$\endgroup$
  • 1
    $\begingroup$ Welcome to Engineering! This looks like a "homework question". In order for such questions to be answered in this site, we need you to add details describing the precise problem you're having. What have you tried to solve this yourself? Please edit your question to include this information. $\endgroup$ – Wasabi Sep 10 '17 at 10:59
  • $\begingroup$ @Wasabi see image file attached above below my question. $\endgroup$ – Surdz Sep 10 '17 at 21:02
1
$\begingroup$

First you have to notice that the contact at A is a roll support, so the reaction force there is perpendicular to the surface. Therefore Fa has 60 degree against horizontal axis. When you draw free body diagram, decompose the Fa reaction force into its horizontal and vertical components:

Fax = Fa Cos60

Fay = Fa Sin60

Then you solve the balance equations of X-axis, Y-axis, and Moments. I would start from the sum of Moments at C equal zero. You will get the value of Fa. Then with the balance equations of X-axis and Y-axis with Fa already known you will get the value of Fcx and Fcy.

I get a slightly different result, probably due to rounding:

Fa = 365.24 N

Fcx = 782.62 N

Fcy = 316.3 N

| improve this answer | |
$\endgroup$
  • $\begingroup$ Mc=0 (xFcy-yFcx)+(xFBy-yFbx)=0 is it like this? $\endgroup$ – Surdz Sep 10 '17 at 4:10
  • $\begingroup$ @Surdz you forgot to include the 300 N loads on your equation, one with 200 mm lever arm, the other one with 400 mm lever arm. $\endgroup$ – JohnDoe Sep 10 '17 at 5:01
  • $\begingroup$ so it will be like; Mc=0 (xFcy-yFcx)+(xFby-yFbx)+(xFay-yFax)=0 [(0*Fcy-0.2(300)]+[(0*Fby-0.4(300)]+[(0.8Fay-0*Fax)]=0 -40-120+0.8Fay=0 -160+0.8Fay=0 Fay=160/0.8=200N can you check why im getting that above and explain how to find Fcx and Fcy $\endgroup$ – Surdz Sep 10 '17 at 6:30
  • 1
    $\begingroup$ It's not very tidy equation, but fundamentally you got things right until you write this [(0.8Fay-0*Fax)]. In this problem, there is Fax, which is Fa Cos60, and that force inflicting moment at point C because when you look at it, Fax has 400 mm lever arm. So that part of the equation should be (+ 0.8*Fay - 0.4*Fax). Then you substitute Fax and Fay with FaCos60 and FaSin60 respectively. You can solve the equation for Fa then. After you get Fa, it will be easy to calculate Fcx and Fcy. $\endgroup$ – JohnDoe Sep 10 '17 at 6:47

Not the answer you're looking for? Browse other questions tagged or ask your own question.